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Have a rectangle bdfg with point a on side bd so that ba equals 16 cm and ad equals 8 cm what fraction of the area of the rectangle is inside triangle baf?
ba=(16/(16+8))bd=(16/24)bd=(2/3)bd area of the rectangle = bd*bf area of triangle = (2/3)bd*bf/2=(1/3)*bd*bf 1/3
Why is a over b plus c over d equal ad plus bc divided by bd always valid?
Because it can be derived from a large number of more basic principles applied to numbers. Since a/b and c/d are defined, then b and d are non-zero. Then a/b = ad/bd since multiplying the numerator and denominator of a fraction by a non-zero number leaves it unchanged. Similarly, c/d = bc/bd [These steps implicitly assume the commutative property of multiplication ie there is no difference between premultiplication and postmultiplication] Another property of numbers: If x = p and y = q then x+y = p+q [equals added to equals result in equals]. So a/b + c+d = ad/bd + bc/bd and finally, using the distributive property of multiplication [by 1/bd] over addition, = (ad + bc)/bd Maybe you wish you hadn't asked!
Does distributive property apply to complex numbers?
Yes, for example (a + bi)(c + di) = ac + adi + bic + bidi, and commutative property works as well --> ac + adi + bci + bdi² --> ac + (ad + bc)i + bd(-1) = (ac - bd) + (ad + bc)i
What does a over b plus c over d equal?
The expression ( \frac{a}{b} + \frac{c}{d} ) can be combined by finding a common denominator. The result is ( \frac{ad + bc}{bd} ). Thus, ( \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} ).
Can AC plus AD plus BC plus BD can be factored as (A plus C)(B plus D)?
Yes, the expression AC + AD + BC + BD can be factored as (A + C)(B + D). This is evident by applying the distributive property, where expanding (A + C)(B + D) yields AB + AD + BC + CD. The terms AC and BD are not present, so the expression can be expressed in a different form, but the original expression itself represents a different factoring structure.
If bd does not equal 0 then a over b plus c over d equals?
If bd ≠ 0, then a/b + c/d (the common denominator is bd) = (a x d)/(b x d) + (c x b)/(d x b) = ad/bd + cb/db = ad/bd + cb/bd = (ad + cb)/ bd
What is the difference between AD and BD?
AD means that Anno Domini , the year of lord and BD means that Before Christ
Is the difference of two rational numbers always a rational number?
Yes. Since they are rational numbers, let's call the first one a/b and the second one c/d where a,b,c, and d are integers. Now we can subtract by finding a common denominator. Let's use bd. So we have ad/bd-cb/bc= (ad-bc)/CD which is rational since we know ad and bc are integers being the product of integers and CD is also an integers. Call ad-bd=P and call CD=Q where P and Q are integers. We now see the difference is of two rationals is rational.
How many different ways can you write a fractions that has a numerator of 2 as a sum of fractions?
Infinitely many ways!Suppose you have the fraction 2/d.Pick any pair of integers a and b where b � 0.Then 2b-ad is and integer, as is bd so that (2b - ad)/bd is a fraction.Consider the fractions a/b and (2b - ad)/bdTheir sum isa/b + (2b-ad)/bd = ad/bd + (2b-ad)/bd = 2b/bd = 2/d - as required.Since a and b were chosen arbitrarily, there are infinitely many possible answers to the question.
Why the sum difference and product of 2 rational numbers rational?
A rational number is one that can be expressed as a/b The sum of two rational numbers is: a/b + c/d =ad/bd + bc/bd =(ad+bc)/bd =e/f which is rational The difference of two rational numbers is: a/b - c/d =ab/bd - bc/bd =(ab-bc)/bd =e/f which is rational The product of two rational numbers is: (a/b)(c/d) =ac/bd =e/f which is rational
Would the difference of a rational number and a rational number be rational?
The difference of two rational numbers is rational. Let the two rational numbers be a/b and c/d, where a, b, c, and d are integers. Any rational number can be represented this way. Their difference is a/b-c/d = ad/bd-cb/bd = (ad-cb)/bd. Products and differences of integers are always integers. This means that ad-cb is an integer, and so is bd. Thus, (ad-cb)/bd is a rational number (since it is the ratio of two integers). This is equivalent to the difference of the original two rational numbers.
Have a rectangle bdfg with point a on side bd so that ba equals 16 cm and ad equals 8 cm what fraction of the area of the rectangle is inside triangle baf?
ba=(16/(16+8))bd=(16/24)bd=(2/3)bd area of the rectangle = bd*bf area of triangle = (2/3)bd*bf/2=(1/3)*bd*bf 1/3
What is the perimeter of the rectangle ABCD when AD equals 3 and BD equals 5?
It is 16 units.
How bad was black Sunday?
Really bad bd bd bd bd bd bd bd bd i dont know bd bd bd bd bd bd bd bd bd bd bd bd bd bd bd bd bd
Why is a over b plus c over d equal ad plus bc divided by bd always valid?
Because it can be derived from a large number of more basic principles applied to numbers. Since a/b and c/d are defined, then b and d are non-zero. Then a/b = ad/bd since multiplying the numerator and denominator of a fraction by a non-zero number leaves it unchanged. Similarly, c/d = bc/bd [These steps implicitly assume the commutative property of multiplication ie there is no difference between premultiplication and postmultiplication] Another property of numbers: If x = p and y = q then x+y = p+q [equals added to equals result in equals]. So a/b + c+d = ad/bd + bc/bd and finally, using the distributive property of multiplication [by 1/bd] over addition, = (ad + bc)/bd Maybe you wish you hadn't asked!
How do two fractions have a difference close to 0?
according to sum of fractions: a/b + c/d = (ad+bc)/bd hence for example: a=c=1 , b=d=2: (ad+bc)/bd = (2+2)/2x2 = 4/4 = 1 sagy
Does distributive property apply to complex numbers?
Yes, for example (a + bi)(c + di) = ac + adi + bic + bidi, and commutative property works as well --> ac + adi + bci + bdi² --> ac + (ad + bc)i + bd(-1) = (ac - bd) + (ad + bc)i
